A178785
a(n) is the smallest n-perfect number of the form 2^(n+1)*L, where L is an odd number with exponents <= n in its prime power factorization, and a(n)=0 if no such n-perfect number exists.
Original entry on oeis.org
60, 6552, 222768, 288288, 87360, 49585536, 25486965504, 203558400, 683289600, 556121548800
Offset: 1
In case of n=2, we have the basis ("2-primes"): 2,3,5,7,8,11,13,... By the formula, we construct from the left m and from the right 2*m. By the condition, m begins from "2-prime" 8. From the right we have 8+1=3^2, therefore from the left we have 8*3^2 and from the right 3^2*(3^3-1)/(3-1)=3^2*13. Thus from the left it should be 8*3^2*13 and from the right 3^2*13*14. Finally, from the left we obtain m=8*3^2*13*7=6552 and from the right we have 2*m=3^2*13*14*8. By the construction, it is the smallest 2-perfect number of the required form. Thus a(2)=6552.
A186887
The smallest infinitary divisor of the n-th infinitary perfect number, which is a perfect square >1.
Original entry on oeis.org
4, 9, 9, 16, 4, 4, 9, 16, 4, 81, 4, 4, 16, 25, 9, 4
Offset: 2
A186889
Oex perfect numbers: n such that A186644(n) = 2*n.
Original entry on oeis.org
6, 18, 20, 100, 1888, 2044928, 33099776, 35021696, 45335936, 533020672
Offset: 1
Let n = 100 with divisors 1, 2, 4, 5, 10, 20, 25, 50, and 100. By the definition in A186643, only 1, 4, 20, 25, 50, 100 among these are oex divisors. Since 1+4+20+25+50+100 = 2*100, 100 is in the sequence.
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for(n=4, 10^9, if(isprime(n), next); d=divisors(n); s=n+1; for(j=2, numdiv(n)-1, for(k=2, 30, if(n%d[j]^k<>0, if(k%2==0, s=s+d[j]); k=30))); if(s==2*n, print(n))) /* Donovan Johnson, Jan 28 2013 */
A361811
Smallest members of infinitary sociable quadruples.
Original entry on oeis.org
1026, 10098, 10260, 41800, 45696, 100980, 241824, 685440, 4938136, 13959680, 14958944, 25581600, 28158165, 32440716, 36072320, 55204500, 74062944, 81128632, 149589440, 178327008, 192793770, 209524210, 283604220, 319848642, 498215416, 581112000, 740629440, 1236402232
Offset: 1
1026 is a term since the iterations of the sum of aliquot infinitary divisors function (A126168) that start with 1026 are cyclic with period 4: 1026, 1374, 1386, 1494, 1026, ..., and 1026 is the smallest member of the quadruple.
The first five quadruples are {1026, 1374, 1386, 1494}, {10098, 15822, 19458, 15102}, {10260, 13740, 13860, 14940}, {41800, 51800, 66760, 83540}, {45696, 101184, 94656, 88944}.
Subsequence of
A004607 (all cycles of length > 2).
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f[p_, e_] := Module[{b = IntegerDigits[e, 2], m}, m = Length[b]; Product[If[b[[j]]>0, 1 + p^(2^(m-j)), 1], {j, 1, m}]]; infs[n_] := Times @@ f @@@ FactorInteger[n] - n; infs[1] = 0; seq[n_] := NestList[infs, n, 4][[2;; 5]] ; q[n_] := Module[{s = seq[n]}, n == Min[s] && Count[s, n] == 1]; Select[Range[10^6], q]
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infs(n) = {my(f = factor(n), b); prod(i=1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], f[i, 1]^(2^(#b-k)) + 1, 1))) - n; }
is(n) = {my(m = n); for(k = 1, 4, m = infs(m); if(k < 4 && m <= n, return(0))); m == n; }
A376889
Numbers k such that A376888(k) = 2*k.
Original entry on oeis.org
6, 60, 90, 336, 5040, 87360, 764400, 11466000, 620568000, 9478560000, 14217840000, 22805874000
Offset: 1
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ff[q_, s_] := (q^(s + 1) - 1)/(q - 1); f[p_, e_] := Module[{k = e, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, AppendTo[s, {p^(m - 1)!, r}];]; m++]; Times @@ ff @@@ s]; fsigma[1] = 1; fsigma[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[10^6], fsigma[#] == 2*# &]
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fdigits(n) = {my(k = n, m = 2, r, s = []); while([k, r] = divrem(k, m); k != 0 || r != 0, s = concat(s, r); m++); s;}
fsigma(n) = {my(f = factor(n), p = f[, 1], e = f[, 2], d); prod(i = 1, #p, prod(j = 1, #d=fdigits(e[i]), (p[i]^(j!*(d[j]+1)) - 1)/(p[i]^j! - 1)));}
is(k) = fsigma(k) == 2*k;
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