A140992
a(0) = 0, a(1) = 1; for n > 1, a(n) = a(n-2) + a(n-1) + A000071(n+1).
Original entry on oeis.org
0, 1, 2, 5, 11, 23, 46, 89, 168, 311, 567, 1021, 1820, 3217, 5646, 9849, 17091, 29523, 50794, 87081, 148820, 253611, 431087, 731065, 1237176, 2089633, 3523226, 5930669, 9968123, 16730831, 28045222, 46954361, 78524160, 131181407
Offset: 0
If n = 4, then a(4) = a(4-2) + a(4-1) + A000071(4+1) = a(2) + a(3) + A000071(5) = 2 + 5 + 4 = 11.
-
LinearRecurrence[{3,-1,-3,1,1},{0,1,2,5,11},40] (* Harvey P. Dale, Jun 12 2014 *)
Corrected (5980669 replaced by 5930669) by
R. J. Mathar, Apr 27 2010
A141019
a(n) is the largest number in the n-th row of triangle A140996.
Original entry on oeis.org
1, 1, 2, 4, 8, 16, 31, 60, 116, 224, 432, 833, 1606, 3096, 5968, 11504, 22175, 42744, 84752, 169880, 340013, 679604, 1356641, 2704954, 5387340, 10718620, 21304973, 42308331, 83945336, 166423276, 329683867, 652627294, 1291020297, 2552209710, 5042305104
Offset: 0
The largest number of 1 is a(0) = 1.
The largest number of 1 1 is a(1) = 1.
The largest number of 1 2 1 is a(2) = 2.
The largest number of 1 4 2 1 is a(3) = 4.
The largest number of 1 8 4 2 1 is a(4) = 8.
The largest number of 1 16 8 4 2 1 is a(5) = 16.
The largest number of 1 31 17 8 4 2 1 is a(6) = 31.
-
A140996 := proc(n,k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; else procname(n-1,k)+procname(n-2,k) +procname(n-3,k)+procname(n-4,k)+procname(n-4,k-1) ; fi; end:
A141019 := proc(n) max(seq(A140996(n,k),k=0..n)) ; end: for n from 0 to 50 do printf("%d,",A141019(n)) ; od: # R. J. Mathar, Sep 19 2008
-
T[n_, k_] := T[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n - 1, 2, k == n-2, 4, k == n-3, 8, True, T[n-1, k] + T[n-2, k] + T[n-3, k] + T[n-4, k] + T[n-4, k-1]];
a[n_] := Table[T[n, k], {k, 0, n}] // Max;
Table[a[n], {n, 0, 34}] (* Jean-François Alcover, Jan 28 2024, after R. J. Mathar *)
Simplified definition and extended by
R. J. Mathar, Sep 19 2008
A141017
List of largest row numbers of Pascal-like triangles with index of asymmetry y = 1 and index of obliqueness z = 0 or z = 1.
Original entry on oeis.org
1, 1, 2, 4, 7, 12, 23, 46, 89, 168, 311, 594, 1194, 2355, 4570, 8745, 16532, 32948, 65761, 129632, 252697, 487647, 936785, 1884892, 3754166, 7407451, 14489982, 28118751, 54868937, 110096666, 219129673, 432847116, 848952949, 1654022768, 3256427202, 6524228863, 12983131874, 25671612977, 50454577444
Offset: 1
Triangle with y = 1 and z = 0 (i.e., triangle A140998) begins as follows:
a(1) = max(1) = 1;
a(2) = max(1, 1) = 1;
a(3) = max(1, 2, 1) = 2;
a(4) = max(1, 4, 2, 1) = 4;
a(5) = max(1, 7, 5, 2, 1) = 7;
a(6) = max(1, 12, 11, 5, 2, 1) = 12;
a(7) = max(1, 20, 23, 12, 5, 2, 1) = 23;
a(8) = max(1, 33, 46, 28, 12, 5, 2, 1) = 46;
a(9) = max(1, 54, 89, 63, 29, 12, 5, 2, 1) = 89;
...
-
# Here, BB is the bivariate g.f. of sequence A140993.
BB := proc(x, y) y*x*(1 - y*x - x^2*y^2 + x^3*y^2)/((1 - x)*(1 - y*x)*(1 - y*x - x^2*y - x^2*y^2)); end proc;
#
# Here, we find the n-th row of sequence A140993 and find the maximum of the row:
ff := proc(n) local xx, k, yy;
xx := 0;
for k from 1 to n do
yy := coeftayl(coeftayl(BB(x, y), x = 0, n), y = 0, k);
xx := max(xx, yy); end do; xx;
end proc;
#
# Here, we print the maxima of the rows:
for i from 1 to 40 do
ff(i);
end do; # Petros Hadjicostas, Jun 10 2019
Comments