cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-15 of 15 results.

A132028 Product{0<=k<=floor(log_4(n)), floor(n/4^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 16, 18, 20, 22, 36, 39, 42, 45, 64, 68, 72, 76, 100, 105, 110, 115, 144, 150, 156, 162, 196, 203, 210, 217, 512, 528, 544, 560, 648, 666, 684, 702, 800, 820, 840, 860, 968, 990, 1012, 1034, 1728, 1764, 1800, 1836, 2028, 2067, 2106, 2145, 2352
Offset: 1

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Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-4 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(26)=floor(26/4^0)*floor(26/4^1)*floor(26/4^2)=26*6*1=156; a(34)=544 since 34=202(base-4) and so
a(34)=202*20*2(base-4)=34*8*2=544.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Formula

Recurrence: a(n)=n*a(floor(n/4)); a(n*4^m)=n^m*4^(m(m+1)/2)*a(n).
a(k*4^m)=k^(m+1)*4^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_4(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_4(n)))/4^((1+floor(log_4(n)))*floor(log_4(n))/2); equality holds for n=k*4^m, 0=0. b(n) can also be written n^(1+floor(log_4(n)))/4^A000217(floor(log_4(n))).
Also: a(n)<=2^(1/4)*n^((1+log_4(n))/2)=1.189207...*4^A000217(log_4(n)), equality holds for n=2*4^m, m>=0.
a(n)>c*b(n), where c=0.4194224417951075977... (see constant A132020).
Also: a(n)>c*2^(1/4)*n^((1+log_4(n))/2)=0.498780...*4^A000217(log_4(n)).
lim inf a(n)/b(n)=0.4194224417951075977..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_4(n))/2)=0.4194224417951075977...*2^(1/4), for n-->oo.
lim sup a(n)/n^((1+log_4(n))/2)=2^(1/4), for n-->oo.
lim inf a(n)/a(n+1)=0.4194224417951075977... for n-->oo (see constant A132020).

A132029 Product{0<=k<=floor(log_5(n)), floor(n/5^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 20, 22, 24, 26, 28, 45, 48, 51, 54, 57, 80, 84, 88, 92, 96, 125, 130, 135, 140, 145, 180, 186, 192, 198, 204, 245, 252, 259, 266, 273, 320, 328, 336, 344, 352, 405, 414, 423, 432, 441, 1000, 1020, 1040, 1060, 1080, 1210, 1232, 1254, 1276
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-5 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(26)=floor(26/5^0)*floor(26/5^1)*floor(26/5^2)=26*5*1=130; a(34)=204 since 34=114(base-5) and so a(34)=114*11*1(base-5)=34*6*1=204.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

  • Mathematica
    Table[Product[Floor[n/5^k],{k,0,Floor[Log[5,n]]}],{n,60}] (* Harvey P. Dale, Oct 16 2019 *)

Formula

Recurrence: a(n)=n*a(floor(n/5)); a(n*5^m)=n^m*5^(m(m+1)/2)*a(n).
a(k*5^m)=k^(m+1)*5^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_5(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_5(n)))/5^((1+floor(log_5(n)))*floor(log_5(n))/2); equality holds for n=k*5^m, 0=0. b(n) can also be written n^(1+floor(log_5(n)))/5^A000217(floor(log_5(n))).
Also: a(n)<=2^((1-log_5(2))/2)*n^((1+log_5(n))/2)=1.2181246...*5^A000217(log_5(n)), equality holds for n=2*5^m, m>=0.
a(n)>c*b(n), where c=0.438796837203638531... (see constant A132021).
Also: a(n)>c*(sqrt(2)/2^log_5(sqrt(2)))*n^((1+log_5(n))/2)=0.534509224...*5^A000217(log_5(n)).
lim inf a(n)/b(n)=0.438796837203638531..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_5(n))/2)=0.438796837203638531...*sqrt(2)/2^log_5(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_5(n))/2)=sqrt(2)/2^log_5(sqrt(2))=1.2181246..., for n-->oo.
lim inf a(n)/a(n+1)=0.438796837203638531... for n-->oo (see constant A132021).

A132030 a(n) = Product_{k=0..floor(log_6(n))} floor(n/6^k), n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 24, 26, 28, 30, 32, 34, 54, 57, 60, 63, 66, 69, 96, 100, 104, 108, 112, 116, 150, 155, 160, 165, 170, 175, 216, 222, 228, 234, 240, 246, 294, 301, 308, 315, 322, 329, 384, 392, 400, 408, 416, 424, 486, 495, 504, 513, 522, 531, 600
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base 6 as n = d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(52) = floor(52/6^0)*floor(52/6^1)*floor(52/6^2) = 52*8*1 = 416;
a(58) = 522 since 58 = 134_6 and so a(58) = 134_6 * 13_6 * 1_6 = 58*9*1 = 522.
		

Crossrefs

For formulas regarding a general parameter p (i.e., terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

  • Maple
    f:= proc(n) option remember; n*procname(floor(n/6)) end proc:
    f(0):= 1:
    seq(f(i),i=1..100); # Robert Israel, Dec 20 2015
  • Mathematica
    Table[Product[Floor[n/6^k], {k, 0, Floor[Log[6, n]]}], {n, 1, 100}] (* G. C. Greubel, Dec 20 2015 *)

Formula

Recurrence: a(n)=n*a(floor(n/6)); a(n*6^m)=n^m*6^(m(m+1)/2)*a(n).
a(k*6^m) = k^(m+1)*6^(m(m+1)/2), for 0
Asymptotic behavior: a(n) = O(n^((1+log_6(n))/2)); this follows from the inequalities below.
a(n) <= b(n), where b(n) = n^(1+floor(log_6(n)))/6^((1+floor(log_6(n)))*floor(log_6(n))/2); equality holds for n=k*6^m, 0=0. b(n) can also be written n^(1+floor(log_6(n)))/6^A000217(floor(log_6(n))).
Also: a(n) <= 2^((1-log_6(2))/2)*n^((1+log_6(n))/2) = 1.236766885...*6^A000217(log_6(n)), equality holds for n=2*6^m and for n=3*6^m, m>=0 (consider 2^((1-log_6(2))/2)=3^((1-log_6(3))/2) since 6=2*3).
a(n) > c*b(n), where c = 0.45071262522603913... (see constant A132022).
Also: a(n) > c*(sqrt(2)/2^log_6(sqrt(2)))*n^((1+log_6(n))/2) = 0.557426449...*6^A000217(log_6(n)).
lim inf a(n)/b(n) = 0.45071262522603913..., for n-->oo.
lim sup a(n)/b(n) = 1, for n-->oo.
lim inf a(n)/n^((1+log_6(n))/2) = 0.45071262522603913...*sqrt(2)/2^log_6(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_6(n))/2) = sqrt(3)/3^log_6(sqrt(3))=1.236766885..., for n-->oo.
lim inf a(n)/a(n+1) = 0.45071262522603913... for n-->oo (see constant A132022).
G.f. g(x) satisfies g(x) = (x+2x^2+3x^3+4x^4+5x^5)*(1 + g(x^6)) + 6*(x^6+x^7+x^8+x^9+x^10+x^11)*g'(x^6). - Robert Israel, Dec 20 2015

A132032 Product{0<=k<=floor(log_8(n)), floor(n/8^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 32, 34, 36, 38, 40, 42, 44, 46, 72, 75, 78, 81, 84, 87, 90, 93, 128, 132, 136, 140, 144, 148, 152, 156, 200, 205, 210, 215, 220, 225, 230, 235, 288, 294, 300, 306, 312, 318, 324, 330, 392, 399, 406, 413, 420, 427, 434
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-8 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(70)=floor(70/8^0)*floor(70/8^1)*floor(70/8^2)=70*8*1=560;
For n=75, 75=113(base-8) and so a(75)=113*11*1(base-8)=75*9*1=675.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Formula

Recurrence: a(n)=n*a(floor(n/8)); a(n*8^m)=n^m*8^(m(m+1)/2)*a(n).
a(k*8^m)=k^(m+1)*8^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_8(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_8(n)))/8^((1+floor(log_8(n)))*floor(log_8(n))/2); equality holds for n=k*8^m, 0=0. b(n) can also be written n^(1+floor(log_8(n)))/8^A000217(floor(log_8(n))).
Also: a(n)<=3^((1-log_8(3))/2)*n^((1+log_8(n))/2) = 1.295758534...*8^A000217(log_8(n)), equality holds for n=3*8^m, m>=0.
a(n)>c*b(n), where c = 0.46456888368647639098... (see constant A132024).
Also: a(n)>c*2^(1/3)*n^((1+log_8(n))/2)=0.4645688836...*1.25992105...*8^A000217(log_8(n)).
lim inf a(n)/b(n)=0.46456888368647639098..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_8(n))/2)=0.46456888368647639098...*2^(1/3), for n-->oo.
lim sup a(n)/n^((1+log_8(n))/2)=sqrt(3)/3^log_8(sqrt(3))=1.295758534..., for n-->oo.
lim inf a(n)/a(n+1)=0.46456888368647639098... for n-->oo (see constant A132024).

A132031 Product{0<=k<=floor(log_7(n)), floor(n/7^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 28, 30, 32, 34, 36, 38, 40, 63, 66, 69, 72, 75, 78, 81, 112, 116, 120, 124, 128, 132, 136, 175, 180, 185, 190, 195, 200, 205, 252, 258, 264, 270, 276, 282, 288, 343, 350, 357, 364, 371, 378, 385, 448, 456, 464, 472, 480, 488
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-7 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(52)=floor(52/7^0)*floor(52/7^1)*floor(52/7^2)=52*7*1=364.
a(58)=464 since 58=112(base-7) and so a(58)=112*11*1(base-7)=58*8*1=464.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

  • Mathematica
    Table[Times@@Floor[n/7^Range[0,Floor[Log[7,n]]]],{n,70}] (* Harvey P. Dale, Oct 11 2017 *)

Formula

Recurrence: a(n)=n*a(floor(n/7)); a(n*7^m)=n^m*7^(m(m+1)/2)*a(n).
a(k*7^m)=k^(m+1)*7^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_7(n))/2)this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_7(n)))/7^((1+floor(log_7(n)))*floor(log_7(n))/2); equality holds for n=k*7^m, 0=0. b(n) can also be written n^(1+floor(log_7(n)))/7^A000217(floor(log_7(n))).
Also: a(n)<=3^((1-log_7(3))/2)*n^((1+log_7(n))/2)=1.270209197...*7^A000217(log_7(n)), equality holds for n=3*7^m, m>=0.
a(n)>c*b(n), where c=0.4587667266997689850200... (see constant A132023).
Also: a(n)>c*(sqrt(2)/2^log_7(sqrt(2)))*n^((1+log_7(n))/2)=0.4587667266...*1.249972544...*7^A000217(log_7(n)).
lim inf a(n)/b(n)=0.4587667266997689850200..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_7(n))/2)=0.4587667266997689850200...*sqrt(2)/2^log_7(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_7(n))/2)=sqrt(3)/3^log_7(sqrt(3))=1.270209197..., for n-->oo.
lim inf a(n)/a(n+1)=0.4587667266997689850200... for n-->oo (see constant A132023).
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