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

Author

Leo Hennig, Apr 09 2025

Keywords

• nonn

Examples

			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
		

Links

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

Crossrefs

Cf. A000396, A155085, A382506.

Programs

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 *)

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