A358929
a(n) is the smallest centered triangular number with exactly n prime factors (counted with multiplicity).
Original entry on oeis.org
1, 19, 4, 316, 136, 760, 64, 4960, 22144, 103360, 27136, 5492224, 1186816, 41414656, 271212544, 559980544, 1334788096, 12943360, 7032930304, 527049293824, 158186536960, 1096295120896, 7871801589760, 154690378792960, 13071965224960, 56262393856, 964655941943296
Offset: 0
a(4) = 136, because 136 is a centered triangular number with 4 prime factors (counted with multiplicity) {2, 2, 2, 17} and this is the smallest such number.
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c[k_] := (3*k^2 + 3*k + 2)/2; a[n_] := Module[{k = 0, ck}, While[PrimeOmega[ck = c[k]] != n, k++]; ck]; Array[a, 18, 0] (* Amiram Eldar, Dec 09 2022 *)
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a(n) = if(n==0, return(1)); for(k=1, oo, my(t=3*k*(k+1)/2 + 1); if(bigomega(t) == n, return(t))); \\ Daniel Suteu, Dec 10 2022
A359235
a(n) is the smallest centered square number with exactly n prime factors (counted with multiplicity).
Original entry on oeis.org
1, 5, 25, 925, 1625, 47125, 2115625, 4330625, 83760625, 1049140625, 6098828125, 224991015625, 3735483578125, 329495166015625, 8193863401953125, 7604781494140625, 216431299462890625, 148146624615478515625, 25926420587158203125, 11071085186929931640625
Offset: 0
a(4) = 1625, because 1625 is a centered square number with 4 prime factors (counted with multiplicity) {5, 5, 5, 13} and this is the smallest such number.
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cs:= n -> 2*n*(n+1)+1:
V:= Vector(12): count:= 0:
for n from 1 while count < 12 do
v:= cs(n);
w:= numtheory:-bigomega(v);
if V[w] = 0 then V[w]:= v; count:= count+1 fi
od:
convert(V,list); # Robert Israel, Dec 22 2022
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bigomega_centered_square_numbers(A, B, n) = A=max(A, 2^n); (f(m, p, n) = my(list=List()); if(n==1, forprime(q=max(p, ceil(A/m)), B\m, if(q%4==1, my(t=m*q); if(issquare(2*t-1), listput(list, t)))), forprime(q=p, sqrtnint(B\m, n), if(q%4==1, my(t=m*q); if(ceil(A/t) <= B\t, list=concat(list, f(t, q, n-1)))))); list); vecsort(Vec(f(1, 2, n)));
a(n) = if(n==0, return(1)); my(x=2^n, y=2*x); while(1, my(v=bigomega_centered_square_numbers(x, y, n)); if(#v >= 1, return(v[1])); x=y+1; y=2*x); \\ Daniel Suteu, Dec 29 2022
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