A372300 Numbers k such that k and k+1 are both primitive infinitary abundant numbers (definition 1, A372298).
812889, 3181815, 20787584, 181480695, 183872535, 307510664, 337206344, 350158808, 523403264, 744074624, 868421504, 1063361144, 1955365125, 2076191864, 2578966215, 3672231255, 4185590408, 5032685384, 7158001304, 8348108535, 10784978295, 16264812135, 20917209495, 24514454055
Offset: 1
Keywords
Crossrefs
Programs
-
PARI
isidiv(d, f) = {my(bne,bde); if (d==1, return (1)); for (k=1, #f~, bne = binary(f[k, 2]); bde = binary(valuation(d, f[k, 1])); if (#bde < #bne, bde = concat(vector(#bne-#bde), bde)); for (j=1, #bne, if (! bne[j] && bde[j], return (0)); ); ); return (1); } idivs(n) = {my(f = factor(n), d = divisors(f), idiv = []); for (k=1, #d, if (isidiv(d[k], f), idiv = concat(idiv, d[k])); ); idiv; } \\ Michel Marcus at A077609 isigma(n) = {my(f = factor(n), b); prod(i=1, #f~, b = binary(f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1)))} ; isab(n) = isigma(n) > 2*n; isprim(n) = select(x -> x
= 2*x, idivs(n)) == []; lista(kmax) = {my(is1 = 0, is2); for(k = 2, kmax, is2 = isab(k); if(is1 && is2, if(isprim(k-1) && isprim(k), print1(k-1, ", "))); is1 = is2);}
Comments