%I #5 Apr 27 2024 03:36:19
%S 812889,3181815,20787584,181480695,183872535,307510664,337206344,
%T 350158808,523403264,744074624,868421504,1063361144,1955365125,
%U 2076191864,2578966215,3672231255,4185590408,5032685384,7158001304,8348108535,10784978295,16264812135,20917209495,24514454055
%N Numbers k such that k and k+1 are both primitive infinitary abundant numbers (definition 1, A372298).
%C The corresponding sequence with definition 2 (A372299) coincides with this sequence for the first 24 terms.
%o (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); }
%o 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
%o 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)))} ;
%o isab(n) = isigma(n) > 2*n;
%o isprim(n) = select(x -> x<n && isigma(x) >= 2*x, idivs(n)) == [];
%o 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);}
%Y Subsequence of A129656, A327635 and A372298.
%Y Cf. A372299.
%Y Similar sequences: A283418, A330872, A361935.
%K nonn
%O 1,1
%A _Amiram Eldar_, Apr 25 2024