A382506 a(n) is the smallest k such that sigma(n) + k is a perfect number.
5, 3, 2, 21, 0, 16, 20, 13, 15, 10, 16, 0, 14, 4, 4, 465, 10, 457, 8, 454, 464, 460, 4, 436, 465, 454, 456, 440, 466, 424, 464, 433, 448, 442, 448, 405, 458, 436, 440, 406, 454, 400, 452, 412, 418, 424, 448, 372, 439, 403, 424, 398, 442, 376, 424, 376, 416, 406, 436, 328, 434, 400
Offset: 1
Keywords
Examples
sigma(1) = 1, 1 + 5 = 6, k = 5. sigma(6) = 12, 12 + 16 = 28, k = 16. sigma(180) = 546, 546 + 7582 = 8128, k = 7582. As sigma(3000) = 9360 and the smallest perfect number at least as large as 9360 is 2^12 * (2^13 - 1) = 33550336 we have a(3000) = 33550336 - sigma(3000) = 33540976. - _David A. Corneth_, Apr 10 2025
Links
- David A. Corneth, Table of n, a(n) for n = 1..10000 (first 3000 terms from Michel Marcus)
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..10000.
Programs
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Mathematica
Do[k=0;s=DivisorSigma[1,n];While[DivisorSigma[1,s+k]!=2*(s+k),k++];a[n]=k,{n,62}];Array[a,62] (* James C. McMahon, Apr 10 2025 *)
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PARI
a(n) = my(s=sigma(n),k=0); while (sigma(s+k) != 2*(s+k), k++); k; \\ Michel Marcus, Mar 30 2025
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PARI
a(n) = {my(s = sigma(n)); forprime(p = 2, oo, my(c = 2^p-1); if(isprime(c) && binomial(c+1, 2) >= s, return(binomial(c+1, 2) - s))) } \\ David A. Corneth, Apr 10 2025
Comments