A337477 Position of the first term >= n in A337475.
1, 4, 17, 570, 63826, 13810338
Offset: 0
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For n = 120, sigma(120) = 360 >= 2*120, thus 120 is not deficient, and we get the next number by applying the prime shift, A003961(120) = 945, and sigma(945) = 1920 >= 945*2, so neither 945 is deficient, so we prime shift once again, and A003961(945) = 9625, which is deficient, as sigma(9625) = 14976 < 2*9625. Thus after two iteration steps we encounter a deficient number, and therefore a(120) = 2.
Array[-1 + Length@ NestWhileList[If[# == 1, 1, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}]] &, #, DivisorSigma[1, #] >= 2 # &] &, 120] (* Michael De Vlieger, Aug 27 2020 *)
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From A003961 A336835(n) = { my(i=0); while(sigma(n) >= (n+n), i++; n = A003961(n)); (i); };
A337473sq(n, k) = if(1==k,k, my(f=factor(k), h = #f~, prevpid=primepi(f[h,1]), e=f[h,2], p, s=1); forstep(i=h-1,0,-1, if(!i,pid=0,pid=primepi(f[i,1])); forstep(j=prevpid,(1+pid),-1, p=prime(j+n); s *= ((p^(1+e)-1)/((p-1)*(p^e)))); if(!pid,return(floor(s))); prevpid = pid; e += f[i,2]); floor(s)); A337474(n) = for(i=0,oo,if(1==A337473sq(i,n),return(i)));
\\ This version uses binary search, which is faster in certain cases: isA337473sq1(n, k) = if(1==k,k, my(f=factor(k), h = #f~, prevpid=primepi(f[h,1]), e=f[h,2], p, s=1); forstep(i=h-1,0,-1, if(!i,pid=0,pid=primepi(f[i,1])); forstep(j=prevpid,(1+pid),-1, p=prime(j+n); s *= ((p^(1+e)-1)/((p-1)*(p^e)))); if(!pid,return(s<2)); prevpid = pid; e += f[i,2]); (s<2)); A337474(n) = if(!bitand(n,n-1),0,my(imin=0,imax=n,imid); for(i=0,oo, imid=(imax+imin)\2; if(1!=isA337473sq1(imid,n), imin = imid+1, if(1!=isA337473sq1(imid-1,n),return(imid), imax = imid-1))));
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