A147794 -id:A147794 - cp's OEIS frontend

A058311 Number of nodes at n-th level in tree in which top node is 1; each node k has children labeled k, k+1, ..., (k+1)^2 at next level.

1, 4, 48, 7918, 463339346, 7134188685100826388, 13246386641449904934758023373599438217628, 643152870463337226096320122089499144560533929707886143570111588898313745804013188842

Offset: 0

N. J. A. Sloane, Dec 09 2000

nonn

Triggered by a comment from Michael Kleber, Dec 08 2009, who said: The algorithm in my paper with Cook lets you compute the equivalent sequence where the children of a node labeled (k) are labeled with all the integers in the interval [p(k), q(k)] where p,q are any polynomials you like (in the paper, p(k)=k+1 and q(k)=2k). For a bunch of p,q the resulting sequence is well known, e.g., p(k)=1, q(k)=k+1 is the Catalan numbers.

M. Cook and M. Kleber, Tournament sequences and Meeussen sequences, Electronic J. Comb. 7 (2000), #R44.

Cf. A008934, A058222, A147780, A147794.

M:=4;
L[0]:=[1]; a[0]:=1;
for n from 1 to M do
L[n]:=[];
t1:=L[n-1];
tc:=nops(t1);
for i from 1 to tc do
t2:=t1[i];
for j from t2 to (t2+1)^2 do
L[n]:=[op(L[n]),j]; od:
a[n]:=nops(L[n]);
#lprint(n,L[n],a[n]);
od:
od:
[seq(a[n],n=0..M)];
# See the reference for a better way to compute this!
p := proc(n,k) option remember; local j ; if n = 1 then k^2+k+2; # (k+1)^2-(k-1) else sum( procname(n-1,j),j=k..(k+1)^2) ; fi; expand(%) ; end proc:
A058311 := proc(n) if n = 0 then 1 ; else subs(k=1, p(n,k)) ; fi; end proc:
for n from 0 do printf("%d,\n", A058311(n)) ; od: # R. J. Mathar, May 04 2009

p[n_, k_] := p[n, k] = If[n == 1, k^2+k+2, Sum[p[n-1, j], {j, k, (k+1)^2}]];
a[n_] := If[n == 0, 1, p[n, 1]];
Table[Print[n, " ", a[n]]; a[n], {n, 0, 7}] (* Jean-François Alcover, Jun 26 2023, after R. J. Mathar *)

Corrected, with Maple program, by N. J. A. Sloane, May 03 2009. Thanks to Max Alekseyev for pointing out that something was wrong.

Replaced a(4), added three more terms - R. J. Mathar, May 04 2009

A147780 Number of nodes at n-th level in tree in which top node is 1; each node k has children labeled 1, 2, ..., (k+1)^2 at next level.

Original entry on oeis.org

1, 4, 54, 8422, 464862602, 7134230598346156958, 13246386641663595526163132113862494582602, 643152870463337226096381089442329605982736165294243832777767297119502149008481206286

Offset: 0

N. J. A. Sloane, May 03 2009

nonn

See the reference in A058311 for a better way to compute this!

A variant of A058311. Cf. A147794.

M:=3;
L[0]:=[1]; a[0]:=1;
for n from 1 to M do
L[n]:=[];
t1:=L[n-1];
tc:=nops(t1);
for i from 1 to tc do
t2:=t1[i];
for j from 1 to (t2+1)^2 do
L[n]:=[op(L[n]),j]; od:
a[n]:=nops(L[n]);
#lprint(n,L[n],a[n]);
od:
od:
[seq(a[n],n=0..M)];
p := proc(n,k) option remember; local j ; if n = 1 then (k+1)^2; else sum( procname(n-1,j),j=1..(k+1)^2) ; fi; expand(%) ; end: A147780 := proc(n) if n = 0 then 1 ; else subs(k=1, p(n,k)) ; fi; end: for n from 0 do printf("%d,\n", A147780(n)) ; od: # R. J. Mathar, May 04 2009

p[n_, k_] := p[n, k] = If[n == 1, (k + 1)^2, Sum[p[n - 1, j], {j, 1, (k + 1)^2}]];
a[n_] := a[n] = If[n == 0, 1, p[n, 1]];
Table[Print[n, " ", a[n]]; a[n], {n, 0, 5}] (* Jean-François Alcover, Nov 28 2023, after R. J. Mathar *)

4 more terms from R. J. Mathar, May 04 2009

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A058311 Number of nodes at n-th level in tree in which top node is 1; each node k has children labeled k, k+1, ..., (k+1)^2 at next level.

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