A213203 -id:A213203 - cp's OEIS frontend

A181482 The sum of the first n integers, with every third integer taken negative.

1, 3, 0, 4, 9, 3, 10, 18, 9, 19, 30, 18, 31, 45, 30, 46, 63, 45, 64, 84, 63, 85, 108, 84, 109, 135, 108, 136, 165, 135, 166, 198, 165, 199, 234, 198, 235, 273, 234, 274, 315, 273, 316, 360, 315, 361, 408, 360, 409, 459, 408, 460, 513, 459, 514, 570, 513, 571, 630

Offset: 1

Jon Perry, Oct 23 2010

The partial sum for the first 10^k terms are 76, 57256, 55722556, 55572225556, 55557222255556,..., i.e., the palindrome 5{k}2{k-1}5{k} plus 1+2*10^(2*k-1). - R. J. Cano, Mar 10 2013, edited by M. F. Hasler, Mar 25 2013

			a(7) = 1 + 2 - 3 + 4 + 5 - 6 + 7 = 10.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Wolfram Alpha, WA Query
Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Cf. A213203, A000217.

a181482 n = a181482_list !! (n-1)
a181482_list = scanl1 (+) $ zipWith (*) [1..] $ cycle [1, 1, -1]
-- Reinhard Zumkeller, Nov 23 2014

c = 0; for (i = 1; i < 100; i++) {c += Math.pow(-1, (i + 1) % 3)*i; document.write(c, ", ");} // Jon Perry, Feb 17 2013

c=0; for (i = 1; i < 100; i++) { c += (1 - (i + 1) % 3 % 2 * 2) * i; document.write(c + ", "); } // Jon Perry, Mar 03 2013

I:=[1,3,0,4,9,3,10]; [n le 7 select I[n] else Self(n-1)+2*Self(n-3)-2*Self(n-4)-Self(n-6)+Self(n-7): n in [1..60]]; // Vincenzo Librandi, Feb 17 2013

a[n_] := Sum[If[Mod[j, 3] == 0, -j, j], {j, 1, n}]; Table[a[i], {i, 1, 50, 1}] (* Jon Perry *)
tri[n_] := n (n + 1)/2; f[n_] := tri@ n - 6 tri@ Floor[n/3]; Array[f, 63] (* Robert G. Wilson v, Oct 24 2010 *)
CoefficientList[Series[-(1 + 2*x + 2*x^3 + x^4 - 3*x^2)/((1 + x + x^2)^2*(x - 1)^3), {x, 0,30}], x] (* Vincenzo Librandi, Feb 17 2013 *)
Table[Sum[k * (-1)^Boole[Mod[k, 3] == 0], {k, n}], {n, 60}] (* Alonso del Arte, Feb 24 2013 *)
With[{nn=20},Accumulate[Times@@@Partition[Riffle[Range[3nn],{1,1,-1}],2]]] (* Harvey P. Dale, Feb 09 2015 *)

a(n)=sum(k=1,n,k*((-1)^(k%3==0)) )  \\ R. J. Cano, Feb 26 2013

a(n)={my(y=n\3);n*(n+1)\2-3*y*(y+1)} \\ R. J. Cano, Feb 28 2013

From R. J. Mathar, Oct 23 2010: (Start)
a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7).
G.f.: -x*(1+2*x+2*x^3+x^4-3*x^2) / ( (1+x+x^2)^2*(x-1)^3 ).
a(n) = 2*A061347(n+1)/9 +4/9 + n*(n+1)/6 + 2*b(n)/3 where b(3k+1) = 0, b(3k) = -3k - 1 and b(3k+2) = 3k + 3. (End)
a(n) = sum((i+1)*A131561(i), i=0..n-1) = A000217(n)-6*A000217(floor(n/3)). [Bruno Berselli, Dec 10 2010]
a(0) = 0, a(n) = a(n-1) + (-1)^((n + 1) mod 3)*n - Jon Perry, Feb 17 2013
a(n) = n*(n+1)/2-3*floor(n/3)*(floor(n/3)+1). - R. J. Cano, Mar 01 2013 [Same as Berselli's formula. - Ed.]
a(3k) = 3k(k-1)/2. - Jon Perry, Mar 01 2013
a(0) = 0, a(n) = a(n-1) + (1 - ((n+1) mod 3 mod 2) * 2) * n. - Jon Perry, Mar 03 2013

More terms added by R. J. Mathar, Oct 23 2010

A222739 Partial sums of the first 10^n terms in A181482.

Original entry on oeis.org

76, 57256, 55722556, 55572225556, 55557222255556, 55555722222555556, 55555572222225555556, 55555557222222255555556, 55555555722222222555555556, 55555555572222222225555555556, 55555555557222222222255555555556, 55555555555722222222222555555555556

Offset: 1

R. J. Cano, Mar 07 2013

Indeed: a(n) is the sum of 2*10^(2n-1)+1 and the palindrome built by repetition of the digits 2 and 5 such that it recalls the number 525.
Let x = 10^n, y = floor(x/3), and B(n) = Sum_{k<=10^n} binomial(floor(k/3),2).
6*B(n) differs from a(n) by (x*(x+1)*(1+(2*x+1)/3))/4-3*y*(3*y+1).

			When n=1, 10^n is 10. By looking at A181482 for its first 10 terms we have the sum: 1+3+0+4+9+3+10+18+9+19, then a(1)=76.

R. J. Cano, Table of n, a(n) for n = 1..49
R. J. Cano, Demonstrative program and additional information.
Index entries for linear recurrences with constant coefficients, signature (1111,-112110,1111000,-1000000).

Cf. A181482, A213203.

repdigit(n,k)=(n!=0)*floor((10/9)*n*10^(k-1));
palindrome(n)=repdigit(5,n)*10^(2*n-1)+repdigit(2,n-1)*10^n+repdigit(5,n);
a(n)=palindrome(n)+(1+2*10^(2*n-1));

Vec(-4*x*(250000*x^3-157875*x^2+6795*x-19)/((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)) + O(x^100)) \\ Colin Barker, Oct 31 2015

a(n) = Sum_{k<=10^n} A181482(k).
From Colin Barker, Oct 31 2015: (Start)
a(n) = 1111*a(n-1)-112110*a(n-2)+1111000*a(n-3)-1000000*a(n-4) for n>4.
G.f.: -4*x*(250000*x^3-157875*x^2+6795*x-19) / ((x-1)*(10*x-1)*(100*x-1)*(1000*x-1)).
(End)

A243106 a(n) = Sum_{k=1..n} (-1)^isprime(k)*10^k.

Original entry on oeis.org

10, -90, -1090, 8910, -91090, 908910, -9091090, 90908910, 1090908910, 11090908910, -88909091090, 911090908910, -9088909091090, 90911090908910, 1090911090908910, 11090911090908910, -88909088909091090, 911090911090908910, -9088909088909091090

Offset: 1

R. J. Cano, Aug 19 2014

Alternative definition: a(n,x)=T(x,1) for a dichromate or Tutte-Whitney polynomial in which the matrix t[i,j] is defined as t[i,j]=Delta(i,j)*((-1)^isprime(i)) and "Delta" is the Kronecker Delta function. - Michel Marcus, Aug 19 2014
If 10 is replaced by 1, then this becomes A097454. If it is replaced by 2, one gets A242002. Choosing powers of the base b=10, as done here, allows one to easily read off the equivalent for any other base b > 4, by simply replacing digits 8,9 with b-2,b-1 (when terms are written in base b). [Comment extended by M. F. Hasler, Aug 20 2014]
There are 2^n ways of taking the partial sum of the first n powers of b=10 if exponent zero is excluded and the signs can be assigned arbitrarily. Conjecture: When expressed in base b, the absolute value for any of these terms only contains digits belonging to {0,1,b-2,b-1}; here {0,1,8,9}.

			n=1 is not prime x^1 = (10)^1 = 10, therefore a(1)=10;
n=2 is prime and x^2 = (10)^2 = 100, taking it negative, a(2) = 10 - 100 = -90;
n=3 also is prime, x^3 = 1000, and we have a(3) = 10 - 100 - 1000 = -1090;
n=4 is not prime, so a(4) = 10 - 100 - 1000 + 10000 = 8910;
n=5 is prime, then a(5) = 10 - 100 - 1000 + 10000 - 100000 = -91090;
Examples of analysis for the concatenation patterns among the terms can be found at the "Additional Information" link.

R. J. Cano, Table of n, a(n) for n = 1..100
R. J. Cano, Additional information.
Eric Weisstein's World of Mathematics, Alternating Series
Eric Weisstein's World of Mathematics, Tutte Polynomial

Cf. A097454.
The same kind of base-independent behavior: A215940, A217626.
Partial sums of alternating series: A181482, A222739, A213203.

Table[Sum[ (-1)^Boole@ PrimeQ@ k*10^k, {k, n}], {n, 19}] (* Michael De Vlieger, Jan 03 2016 *)

ap(n, x)={my(s); forprime(p=1, n, s+=x^p); s}
a=(n, x=10)->(x^(n+1)-1)/(x-1)-2*ap(n, x)-1;

Delta=(i, j)->(i==j); /* Kronecker's Delta function */
t=n->matrix(n, n, i, j, Delta(i, j)*((-1)^isprime(i))); /* coeffs t[i, j] */
/* Tutte polynomial over n */
T(n, x, y)={my(t0=t(n)); sum(i=1, n, sum(j=1, n, t0[i, j]*(x^i)*(y^j)))};
a=(n, x=10)->T(n, x, 1);

A243106(n,b=10)=sum(k=1,n,(-1)^isprime(k)*b^k) \\ M. F. Hasler, Aug 20 2014

a(n,x) = Sum_{k=1..n} (-1)^isprime(k)*(x^k), for x=10 in decimal.

Definition simplified by N. J. A. Sloane, Aug 19 2014

Definition further simplified and more terms from M. F. Hasler, Aug 20 2014

cp's OEIS Frontend

A181482 The sum of the first n integers, with every third integer taken negative.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

JavaScript

JavaScript

Magma

Mathematica

PARI

PARI

Formula

Extensions

A222739 Partial sums of the first 10^n terms in A181482.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A243106 a(n) = Sum_{k=1..n} (-1)^isprime(k)*10^k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

Extensions