A003590 -id:A003590 - cp's OEIS frontend

A096884 a(n) = 101^n.

1, 101, 10201, 1030301, 104060401, 10510100501, 1061520150601, 107213535210701, 10828567056280801, 1093685272684360901, 110462212541120451001, 11156683466653165551101, 1126825030131969720661201, 113809328043328941786781301, 11494742132376223120464911401, 1160968955369998535166956051501

Offset: 0

Paul Barry, Jul 14 2004

A185817(n) = smallest m such that in decimal representation n is a prefix of a(m).
a(n) gives the n-th row of Pascals' triangle (A007318) as long as all the binomial coefficients have at most two digits, otherwise the binomial coefficients with more than two digits overlap. - Daniel Forgues, Aug 12 2012
From Peter M. Chema, Apr 10 2016: (Start)
One percent growth applied n times increases a value by factor of a(n)/10^(2n), since 1% increases using "1.01". Therefore (a(n)/10^(2n) - 1)*100 = the percentage increase of one percent growth applied n times.
For instance, 432 increasing by 1% three times gives 445.090032 (i.e., 432*1.01^3), which is 1.030301 (a(3)/10^(2*3)) times 432 or a 3.0301% increase from the original 432 ((a(3)/10^(2*3)-1)*100 = 3.0301). (End)

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (101).

Cf. A007318, A003590, A001020, A097659.

Table[101^n, {n, 0, 15}] (* Ilya Gutkovskiy, Apr 10 2016 *)

a(n)=101^n \\ Charles R Greathouse IV, Oct 16 2015

my(x='x+O('x^20)); Vec(1/(1-101*x)) \\ Altug Alkan, Apr 10 2016

a(n) = Sum_{k=0..n} binomial(n, k)*10^(n-k).
a(n) = A096883(2n).
a(n) = 101^n. a(n) = Sum_{k=0..n,} binomial(n, k)*100^k. - Paul Barry, Aug 24 2004
G.f.: 1/(1-101*x). - Philippe Deléham, Nov 25 2008
E.g.f.: exp(101*x). - Ilya Gutkovskiy, Apr 10 2016

A133342 Concatenation of binary expansion of n-th row of Pascal's triangle.

Original entry on oeis.org

1, 11, 1101, 111111, 11001101001, 1101101010101011, 111011111010011111101, 111110101100011100011101011111, 110001110011100010001101110001110010001, 11001100100101010011111101111110101010010010010011

Offset: 0

Jonathan Vos Post, Oct 20 2007

Binary analog of A003590. More generally, this sequence is the 2nd row of the matrix whose k-th row is the concatenation of the base-k expression of n-th row of Pascal's triangle. The 10th row of that array is A003590.

a(n) is a repunit for n = 0, 1, 3, but otherwise not since C(n,1) = n has a 0-bit apart from at n=2^k-1 and there C(n,2) = n*(n-1)/2 == 1 (mod 4) has a 0-bit. - Kevin Ryde and Bernard Schott, Nov 11 2021

			a(0) = 1 because the 0th row of Pascal's triangle is 1.
a(1) = 11 because the 1st row of Pascal's triangle is 1,1 which concatenates to 11.
a(2) = 1101 because the 2nd row of Pascal's triangle is 1,2,1 which in binary is 1,10,1 which concatenates to 1101.
a(3) = 111111 because the 3rd row of Pascal's triangle is 1,3,3,1 which in binary is 1,11,11,1 which concatenates to 111111.
a(4) = 110010101001 because the 4th row of Pascal's triangle is 1,4,6,4,1 which in binary is 1,100,110,100,1 which concatenates to 11001101001.
a(5) = 1101101010101011 because the 5th row of Pascal's triangle is 1,5,10,10,5,1 which in binary is 1,101,1010,1010,101,1 which concatenates to 1101101010101011.
a(6) = 111011111010011111101 because the 6th row of Pascal's triangle is 1,6,15,20,15,6,1 which in binary is 1,110,1111,10100,1111,110,1 which concatenates to 111011111010011111101.
The array of base k concatenations begins:
  k/n  0  1   2    3        4
  1.|  1  11  1111 11111111 1111111111111111  2^(n-1) repetitions of 1
  2.|  1  11  1101 111111   11001101001
  3.|  1  11  121  110101   11120111
  4.|  1  11  121  1331     11012101
  5.|  1  11  121  1331     141141
  6.|  1  11  121  1331     141041

Harvey P. Dale, Table of n, a(n) for n = 0..38

Cf. A000217, A002275, A003590, A007318, A349304.

catL := proc(L) local resul,a ; resul:=0 ; for a in L do resul := resul*10^(max(ilog10(a)+1,1))+a ; od: RETURN(resul) ; end: A133342 := proc(n) local prow,k ; prow := [1] ; for k from 1 to n do prow := [op(prow), convert(binomial(n,k),binary) ] ; od: catL(prow) ; end: seq(A133342(n),n=0..11) ; # R. J. Mathar, Jan 08 2008

FromDigits[Flatten[IntegerDigits[#,2]]]&/@Table[Binomial[n,k],{n,0,10},{k,0,n}] (* Harvey P. Dale, Apr 12 2020 *)

a(n) = Concatenate[k=1..n] binomial(n, k) (base 2).

a(n) = Concatenate[i=A000217(n)..A000217(n+1)] A007088(A007318(i)).

Corrected and extended by R. J. Mathar, Jan 08 2008

A349304 Terms of sequence A133342 interpreted as numbers written in base 2 and converted here to base 10.

Original entry on oeis.org

1, 3, 13, 63, 1641, 55979, 1963261, 1051838303, 427823653777, 899765549835411, 962612860717614517, 16451240378526759988983, 275851420933163891499583577, 75079775107059278497058879826395, 1275913767034795679914048924532788093, 1386272923043819758818080175342943182028095

Offset: 0

Bernard Schott, Nov 14 2021

a(n) is a Mersenne number (A000225) iff n = 0, 1, 3 (see proof in A133342).

Cf. A000225, A003590, A007318, A133342.

a[n_] := FromDigits[Join @@ Table[IntegerDigits[Binomial[n, k], 2], {k, 0, n}], 2]; Array[a, 15, 0] (* Amiram Eldar, Nov 14 2021 *)

a(n) = fromdigits(concat(vector(n+1, k, binary(binomial(n, k-1)))), 2); \\ Michel Marcus, Nov 14 2021

A369152 Total number of digits in row n of Pascal's triangle.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 10, 12, 14, 18, 25, 28, 31, 38, 44, 50, 57, 64, 69, 76, 88, 96, 102, 114, 125, 134, 142, 154, 166, 178, 191, 202, 215, 230, 244, 256, 268, 288, 303, 316, 334, 356, 372, 388, 411, 428, 447, 470, 490, 506, 529, 554, 573, 590, 618, 642, 660, 686

Offset: 0

Bartlomiej Pawlik, Jan 14 2024

Cf. A007318, A055642, A003590.

a:= n-> length(cat(seq(binomial(n,k), k=0..n))):
seq(a(n), n=0..57);  # Alois P. Heinz, Jan 15 2024

A369152[n_Integer] := Total[IntegerLength[Binomial[n, #]] & /@ Range[0, n]]
First50Terms = Table[A369152[n], {n, 0, 49}]

a(n) = #concat(vector(n+1, k, Str(binomial(n,k-1)))); \\ Michel Marcus, Jan 18 2024

from math import comb
def A369152(n): return sum(len(str(comb(n,k))) for k in range(n+1)) # Chai Wah Wu, Feb 15 2024

a(n) = 1 + n + Sum_{k=0..n} floor(log_10(C(n,k))).

a(n) = A055642(A003590(n)). - Michel Marcus, Jan 15 2024

cp's OEIS Frontend

A096884 a(n) = 101^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A133342 Concatenation of binary expansion of n-th row of Pascal's triangle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions

A349304 Terms of sequence A133342 interpreted as numbers written in base 2 and converted here to base 10.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

A369152 Total number of digits in row n of Pascal's triangle.

Views

Author

Keywords

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula