25431=213+53+47+36+1
25432=213+53+47+36+2
25433=213+53+47+36+3
25434=213+53+47+36+4
25435=213+53+47+36+5
25436=213+53+47+36+6
25437=213+53+47+36+7
25438=213+53+47+36+8
25439=213+53+47+36+9
The occurrences of such decades are relatively frequent. There are 600 d-powerful numbers in the range from 1 to 17463 inclusive. This represents a relative frequency of a little over 3.4%. D-powerful numbers can be written in more than one way. For example, 25432 can be written in five different ways:
25432=213+53+47+36+2
25432=212+54+45+39+22
25432=210+54+46+39+22
25432=211+55+43+39+29
25432=212+54+4+39+210
Similarly, 25437 can be written in two ways:
25437=213+53+47+36+7
25437=211+5+42+38+75
Finally, 25439 can be written in two ways:
25439=213+53+47+36+9
25439=213+53+47+32+33
OEIS A007532 refers to d-powerful numbers as simply powerful numbers and lists the following initial member of the sequence:
1, 2, 3, 4, 5, 6, 7, 8, 9, 24, 43, 63, 89, 132, 135, 153, 175, 209, 224, 226, 262, 264, 267, 283, 332, 333, 334, 357, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 407, 445, 463, 518, 598, 629, 739, 794, 849, 935, 994, 1034So I have another nine d-powerful days to go. This website gives some examples of interesting d-powerful numbers with additional special properties. For example, consider the case of d-powerful palindromic primes. There is only one such 3 digit number, 373, and it has two representations:
373=31+73+33
373=34+72+35
There is only one 5 digit d-powerful palindromic prime and that is 98389:
98389=94+81+31+85+95
Another category are d-powerful numbers that a palindromic numbers and have a representation where the exponents are symmetric. The first such numbers are:
262=27+61+27
4224=43+211+211+43
39393=39+91+32+91+39
79597=71+93+57+93+71
The smallest d-powerful, pandigital number is 1023456879 with representation:
1023456879=11+01+227+318+46+51+611+89+75+97
Here is the SageMath code to determine the indices of a number if it is d-powerful (25430 is used as the test number but any number can be substituted for it, provided it has five digits:
xxxxxxxxxx
# since 2^15 = 32768 this method works for 5 digit numbers up a little below this number
number=25430
N=number.digits()
N.reverse()
a,b,c,d,e=var('a,b,c,d,e')
for a in [1..15]:
for b in [1..15]:
for c in [1..15]:
for d in [1..15]:
for e in [1..15]:
if N[0]^a+N[1]^b+N[2]^c+N[3]^d+N[4]^e==number:
print number,"-->",a,b,c,d,e
break
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