A132028 Product{0<=k<=floor(log_4(n)), floor(n/4^k)}, n>=1.
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
Keywords
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
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.
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
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.
Links
- Harvey P. Dale, Table of n, a(n) for n = 1..1000
Crossrefs
Programs
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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.
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
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.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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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
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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.
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
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
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.
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
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
Programs
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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).
Comments