0,2

Because of the identity linking triangular numbers T(n) and pentagonal numbers P(n), namely: P(n)= [T(3n-1)]/3, we may write the Iterated pentagonal numbers in terms of triangular numbers: P(2,n) = (1/6){[T(3n-1)] 2} - [(1/6)T(3n-1)]; P(3,n) = (1/24)[T(3n-1)] 4 - (1/12)[T(3n-1)] 3 - (1/24)[T(3n-1)] 2 - (1/12)T(3n-1); and so forth.

J. V. Post, "Iterated Polygonal Numbers", preprint.

J. V. Post, "Iterated Triangular Numbers", preprint.

Eric Weisstein's World of Mathematics, "Pentagonal Number."

Where P(n) = n*(3*n-1)/2 = the n-th pentagonal number, we define: a(1, n) = P(n) = n*(3*n-1)/2 = (1/2)*[3*(n^2)-n]; a(2, n) = P(P(n)) = (1/2)*{(1/2)*[3*(n^2)-n]}*{(3/2)[3*(n^2)-n] - 1} = (27/8)*(n^4) - (9/4)*(n^3)- (3/2)*(n^2) + (1/4)*n; a(3, n) = P(P(P(n))) = (2187/128)*(n8) - (729/32)*(n^7)- (243/64)*(n^6) + (81/16)*(n^5) - (621/128)*(n^4) + (63/32)*(n^3) + (15/32)*(n^2) - (1/4)*n; P(4, n) = P(P(P(P(n)))) = (14348907/32768)*(n^16) - (4782969/4096)*(n^15)- (1594323/4096)*(n^14) + (4782969/4096)*(n^13) - (621/128)*(n^12) + (295245/1024)*(n^11) + (1161297/5096)*(n^10) - (32805/512)*(n^9) - (32805/1024)*(n^8) + (64881/1024)*(n^7) + (1215/256)*(n^6) - (2025/128)*(n^5) - (405/2048)*(n^4) + (135/256)*(n^3) - (15/32)*(n^2) + (1/16)*n, and in general: a(k+1, n) = P(a(k, n), n) = a(k, n)*(3*a(k, n)-1)/2.

a(3) = 1820 because a(1) = the first pentagonal number = 5, a(2) = the 5th pentagonal number = 35, and a(3) = the 35th pentagonal number = 1820.

Cf. A007501, A000326.

Sequence in context: A001802 A122590 A057991 this_sequence A000871 A035416 A042029

Adjacent sequences: A099134 A099135 A099136 this_sequence A099138 A099139 A099140

easy,nonn,uned

Jonathan Vos Post (jvospost2(AT)yahoo.com), Nov 14 2004