A382929 Smallest number k such that k + n + sigma(n) is a perfect number.

4, 1, 21, 17, 17, 10, 13, 5, 6, 0, 5, 456, 1, 458, 457, 449, 461, 439, 457, 434, 443, 438, 449, 412, 440, 428, 429, 412, 437, 394, 433, 401, 415, 408, 413, 369, 421, 398, 401, 366, 413, 358, 409, 368, 373, 378, 401, 324, 390, 353, 373, 346, 389, 322, 369, 320, 359, 348, 377, 268

Offset

1,1

Links

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 2500 terms from Michel Marcus)

Example

a(10) = 0, because 10 + sigma(10) = 28, which is perfect.
a(12) = 456, because 456 + 12 + sigma(12) = 496, which is perfect.
As 496 is the smallest perfect number at least as large as sigma(60) + 60 = 168 + 60 = 228 we have a(60) = 496 - 228 = 268. - David A. Corneth, Apr 10 2025

Mathematica

Do[k=0; s=DivisorSigma[1, n]; While[DivisorSigma[1, s+n+k]!=2*(s+n+k), k++]; a[n]=k, {n, 60}]; Array[a, 60] (* James C. McMahon, Apr 10 2025 *)

Prog

(PARI) a(n) = my(k=0); while (sigma(k+n+sigma(n)) != 2*(k+n+sigma(n)), k++); k; \\ Michel Marcus, Apr 09 2025
(PARI) a(n) = {my(s = sigma(n) + 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
(PARI) a(n) = my(v = [6, 28, 496, 8128, 33550336, 8589869056], x=n+sigma(n), k=0); for (i=1, #v-1, if ((x > v[i]) && (x <= v[i+1]), k = i; break)); v[k+1] - x; \\ Michel Marcus, Apr 11 2025

Crossrefs

Cf. A000396, A155085, A382506.

Sequence in context: A159841 A202550 A364760 * A142472 A360089 A299445

Adjacent sequences: A382926 A382927 A382928 * A382930 A382931 A382932

Keyword

nonn

Author

Leo Hennig, Apr 09 2025

Status

approved