A017281 -id:A017281 - cp's OEIS frontend

A169669 (first digit of n) * (last digit of n) in decimal representation.

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 0

Offset: 0

Reinhard Zumkeller, Apr 05 2010

a(n) = A000030(n)*A010879(n);
a(n) = A115300(n) for n<=100, A115300(101) = 0;
a(n) = A111707(n) for n<=109, A111707(110) = 1;
0 <= a(n) <= 81, range = A174995;
a(10*n + n mod 10) = a(n);
a(A008592(n)) = 0;
A000030(a(A017281(n))) = A000030(A017281(n));
a(n) = a(A004086(n))*A168184(n);

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A088133.

Cf. A000030, A008592, A010879, A111707, A115300, A174995.

a169669 n = a000030 n * mod n 10
-- Reinhard Zumkeller, Apr 29 2015

def a(n): return int(str(n)[0])*(n%10)
print([a(n) for n in range(81)]) # Michael S. Branicky, Jul 13 2022

A306277 Numbers congruent to 1 or 8 mod 10.

Original entry on oeis.org

1, 8, 11, 18, 21, 28, 31, 38, 41, 48, 51, 58, 61, 68, 71, 78, 81, 88, 91, 98, 101, 108, 111, 118, 121, 128, 131, 138, 141, 148, 151, 158, 161, 168, 171, 178, 181, 188, 191, 198, 201, 208, 211, 218, 221, 228, 231, 238, 241, 248, 251, 258, 261, 268, 271, 278, 281, 288, 291, 298, 301, 308, 311, 318, 321

Offset: 1

Davis Smith, Feb 02 2019

A007310(a(n)+1) is always a multiple of 5.
a(1) = 1, a(n+1) = a(n)+7 when n is odd, a(n+1) = a(n)+3 when n is even.
a(n) mod 6 follows the following pattern: 1,2,5,0,3,4,1,2,5,0,3,4, and so on.
A020639(A007310(a(n)+1)) = 5.

Davis Smith, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001622, A020639, A007310, A000034.
Cf. A017281, A017365 (bisections).
One less than A273669.

Maple
```
seq(seq(10*i+j, j=[1, 8]), i=0..350);
```

Select[Range[350], MemberQ[{1, 8}, Mod[#, 10]] &]

for(n=1, 350, if((n%10==1) || (n%10==8), print1(n, ", ")))

Vec(x*(1 + 7*x + 2*x^2) / ((1 - x)^2*(1 + x)) + O(x^40)) \\ Colin Barker, Feb 09 2019

a(n) = 5*n - 2*A000034(n+1).
a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.
a(n) = A273669(n) - 1. - Antti Karttunen, Feb 07 2019
G.f.: x*(1 + 7*x + 2*x^2) / ((1 - x)^2*(1 + x)). - Colin Barker, Feb 09 2019
E.g.f.: 2 + (5*x - 3)*exp(x) + exp(-x). - David Lovler, Sep 07 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = (5+sqrt(5))^(3/2)*phi*Pi/(100*sqrt(2)) + log(phi)/(2*sqrt(5)) + log(2)/5, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023

A348487 Positive numbers whose square starts and ends with exactly one 1.

Original entry on oeis.org

1, 11, 39, 41, 101, 111, 119, 121, 129, 131, 139, 141, 319, 321, 329, 331, 349, 351, 359, 361, 369, 371, 379, 381, 389, 391, 399, 401, 409, 411, 419, 421, 429, 431, 439, 441, 1001, 1009, 1011, 1019, 1021, 1029, 1031, 1039, 1041, 1099, 1101, 1109, 1111, 1119, 1121, 1129, 1131, 1139

Offset: 1

Bernard Schott, Oct 21 2021

When a square ends with 1, this square ends with exactly one 1.

Sequences A000533 and A253213 show that there are an infinity of terms. The square of their terms, for n >= 3, starts and ends with exactly one 1. Also, the numbers 119, 1119, 11119, ..., ((10^k + 71) / 9)^2, (k >= 3) are terms. The squares ((10^k + 71) / 9)^2, have the last digit 1 and because 12*10^(2*k - 3) < ((10^k + 71) / 9)^2 <13*10^(2*k - 3), for k >= 3, the squares ((10^k + 71) / 9)^2, k >= 4, start with 12. - Marius A. Burtea, Oct 21 2021

			39 is a term since 39^2 = 1521.
109 is not a term since 109^2 = 11881.
119 is a term since 119^2 = 14161.

Cf. A045855, A090771, A253213, A273372 (squares ending with 1), A017281, A017377.
Cf. A000533, A253213 for n >= 2 (subsequences).
Subsequence of A305719.

[1] cat [n:n in [2..1200]|Intseq(n*n)[1] eq 1 and Intseq(n*n)[#Intseq(n*n)] eq 1 and Intseq(n*n)[-1+#Intseq(n*n)] ne 1]; // Marius A. Burtea, Oct 21 2021

Join[{1}, Select[Range[11, 1200], (d = IntegerDigits[#^2])[[1]] == d[[-1]] == 1 && d[[2]] != 1 &]] (* Amiram Eldar, Oct 21 2021 *)

isok(k) = my(d=digits(sqr(k))); (d[1]==1) && (d[#d]==1) && if (#d>2, (d[2]!=1) && (d[#d-1]!=1), 1); \\ Michel Marcus, Oct 21 2021

from itertools import count, takewhile
def ok(n):
  s = str(n*n); return len(s.rstrip("1")) == len(s.lstrip("1")) == len(s)-1
def aupto(N):
  r = takewhile(lambda x: x<=N, (10*i+d for i in count(0) for d in [1, 9]))
  return [k for k in r if ok(k)]
print(aupto(1140)) # Michael S. Branicky, Oct 21 2021

A154428 Primes of the form 50n^2 + 10n + 1.

Original entry on oeis.org

61, 1301, 1861, 2521, 5101, 7321, 8581, 9941, 14621, 16381, 20201, 24421, 26681, 34061, 36721, 51521, 68821, 76441, 97241, 101701, 106261, 110921, 135721, 163021, 168781, 199081, 205441, 218461, 252761, 282001, 304981, 312841, 337021, 353641

Offset: 1

Vincenzo Librandi, Jan 09 2009

Subsequence of A027862 associated with the values of A027861 that are multiples of 5. [R. J. Mathar, Jan 12 2009]

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A017281, A048161.

Filtered(List([1..100],n->50*n^2+10*n+1),IsPrime); # Muniru A Asiru, Apr 25 2019

[a: n in [0..100] | IsPrime(a) where a is 50*n^2 + 10*n + 1]; // Vincenzo Librandi, Jul 23 2012

select(isprime,[50*n^2+10*n+1$n=1..100])[]; # Muniru A Asiru, Apr 25 2019

Select[Table[50n^2+10n+1,{n,0,200}],PrimeQ] (* Vincenzo Librandi, Jul 23 2012 *)

for (n=0, 100, if (isprime (k=50*n^2+10*n+1), print1 (k, ", "))); \\ Vincenzo Librandi, Jul 23 2012

Replaced 13721 by 135721 - R. J. Mathar, Jan 12 2009

A168459 a(n) = (10n + 5(-1)^n - 3)/2.

Original entry on oeis.org

1, 11, 11, 21, 21, 31, 31, 41, 41, 51, 51, 61, 61, 71, 71, 81, 81, 91, 91, 101, 101, 111, 111, 121, 121, 131, 131, 141, 141, 151, 151, 161, 161, 171, 171, 181, 181, 191, 191, 201, 201, 211, 211, 221, 221, 231, 231, 241, 241, 251, 251, 261, 261, 271, 271, 281

Offset: 1

Vincenzo Librandi, Nov 26 2009

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A017281, A168457, A168282.

[1+10*Floor(n/2): n in [1..70]]; // Vincenzo Librandi, Sep 19 2013

Table[5 n + 5 (-1)^n/2 - 3/2, {n, 60}] (* Bruno Berselli, Sep 16 2013 *)
Table[1 + 10 Floor[n/2], {n, 70}] (* or *) CoefficientList[Series[(1 + 10 x - x^2)/((1 + x) (x - 1)^2), {x, 0, 70}], x] (* Vincenzo Librandi, Sep 19 2013 *)

a(n) = 10*n - a(n-1) - 8 with n>1, a(1)=1.
From Bruno Berselli, Sep 16 2013: (Start)
G.f.: x*(1 + 10*x - x^2)/((1+x)*(1-x)^2).
a(n) = A168457(n) - 1 = 2*A168282(n) - 1.
a(n) = a(n-1) +a(n-2) -a(n-3). (End)
a(n) = 1 + 10*floor(n/2). - Vincenzo Librandi, Sep 19 2013
E.g.f.: (1/2)*(5 - 2*exp(x) + (10*x - 3)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 23 2016

New definition by Bruno Berselli, Sep 16 2013

A208260 Nonprime numbers starting and ending with digit 1.

Original entry on oeis.org

1, 111, 121, 141, 161, 171, 1001, 1011, 1041, 1071, 1081, 1101, 1111, 1121, 1131, 1141, 1161, 1191, 1211, 1221, 1241, 1251, 1261, 1271, 1281, 1311, 1331, 1341, 1351, 1371, 1391, 1401, 1411, 1421, 1431, 1441, 1461, 1491, 1501, 1521, 1541, 1551, 1561, 1581, 1591

Offset: 1

Jaroslav Krizek, Feb 24 2012

Complement of A062332 with respect to A208259. Supersequence of A208261 (nonprime numbers with all divisors starting and ending with digit 1).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A208259 (number starting and ending with a number 1), A062332 (primes starting and ending with a number 1).

Cf. A010051, A017281, A131835, A000030.

a208260 n = a208260_list !! (n-1)
a208260_list = filter ((== 0) . a010051') a208259_list
-- Reinhard Zumkeller, Jul 16 2014

Select[Range[2000], ! PrimeQ[#] && First[IntegerDigits[#]] == 1 && Last[IntegerDigits[#]] == 1 &] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)
Join[{1},Select[Range[2000],CompositeQ[#]&&NumberDigit[#,0] == NumberDigit[ #,IntegerLength[ #]-1]==1&]] (* Harvey P. Dale, Aug 01 2021 *)

(1 - A010051(a(n))) * A000030(a(n)) * (a(n) mod 10) = 1. - Reinhard Zumkeller, Jul 16 2014

A273372 Squares ending in digit 1.

Original entry on oeis.org

1, 81, 121, 361, 441, 841, 961, 1521, 1681, 2401, 2601, 3481, 3721, 4761, 5041, 6241, 6561, 7921, 8281, 9801, 10201, 11881, 12321, 14161, 14641, 16641, 17161, 19321, 19881, 22201, 22801, 25281, 25921, 28561, 29241, 32041, 32761, 35721, 36481, 39601, 40401

Offset: 1

Vincenzo Librandi, May 21 2016

Intersection of A000290 and A017281; also, union of A017282 and A017378. The square roots are in A017281 or in A017377 (numbers ending in 1 or 9, respectively). - David A. Corneth, May 22 2016

Seiichi Manyama, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000290, A090771, A132356.
Cf. A017281 (numbers ending in 1), A017283 (cubes ending in 1).
Cf. similar sequences listed in A273373.

/* By definition: */ [n^2: n in [0..200] | Modexp(n,2,10) eq 1];

[5*(10*n+(-1)^n-5)*(2*n+(-1)^n-1)/4+1: n in [1..50]];

Table[5 (10 n + (-1)^n - 5) (2 n + (-1)^n - 1)/4 + 1, {n, 1, 50}]

A273372_list = [(10*n+m)**2 for n in range(10**3) for m in (1,9)] # Chai Wah Wu, May 24 2016

p (1..(n + 1) / 2).inject([]){|s, i| s + [(10 * i - 9) ** 2, (10 * i - 1) ** 2]}[0..n - 1] # Seiichi Manyama, May 24 2016

G.f.: x*(1 + 80*x + 38*x^2 + 80*x^3 + x^4) / ((1 + x)^2*(1 - x)^3).
a(n) = 10*A132356(n-1) + 1 = 5*(10*n+(-1)^n-5)*(2*n+(-1)^n-1)/4+1.
a(n) = (5*n - 5/2 + (3/2)*(-1)^n)^2 = 25*n^2 - 25*n + 17/2 + 15*n*(-1)^n - (15/2)*(-1)^n. - David A. Corneth, May 21 2016
a(n) = A090771(n)^2. - Michel Marcus, May 25 2016
Sum_{n>=1} 1/a(n) = Pi^2*(3+sqrt(5))/50. - Amiram Eldar, Feb 16 2023

Edited by Bruno Berselli, May 24 2016

A306289 The smallest prime factor of numbers greater than 1 and coprime to 6.

Original entry on oeis.org

5, 7, 11, 13, 17, 19, 23, 5, 29, 31, 5, 37, 41, 43, 47, 7, 53, 5, 59, 61, 5, 67, 71, 73, 7, 79, 83, 5, 89, 7, 5, 97, 101, 103, 107, 109, 113, 5, 7, 11, 5, 127, 131, 7, 137, 139, 11, 5, 149, 151, 5, 157, 7, 163, 167, 13, 173, 5, 179, 181, 5, 11, 191, 193

Offset: 1

Davis Smith, Feb 03 2019

nonn

a(n) is the least prime factor of the n-th number that is greater than 1 and congruent to 1 or 5 (mod 6).
a(n) = 5 when n is congruent to {1, 8} (mod 10) (n is a term in A017281, A017365, or A306277). a(n) = 7 when n is congruent to {2, 11} (mod 14) but not {1, 8} (mod 10). a(n) = 11 when n is congruent to {3, 18} (mod 22) but not a case where it equals 5 or 7. a(n) = 13 when n is congruent to {4, 21} (mod 26) (n is a term in A306285) but not a case where it equals 5, 7, or 11. a(n) = 17 when n is congruent to {5, 28} (mod 34) but not a case where it equals 5, 7, 11, or 13. a(n) = 19 when n is congruent to {6, 31} (mod 38) (n is a term in A306331) but not a case where it equals 5, 7, 11, 13, or 17.
Conjecture: This pattern continues indefinitely. a(n) = A007310(m + 1) when n is congruent to {m, A306277(m + 1)} (mod A091999(m + 1)) but not congruent to {k, A306277(k + 1)} (mod A091999(k + 1)), m > k >= 1. The indices of the first appearance of a number in this sequence supports this conjecture in that they are never, for m > 0, congruent to A306277(m + 1) mod A091999(m + 1).

			a(n) is the least term, other than 0, in n-th row of the array A(m,n), where A(m,n) is A007310(m + 1) when A007310(n + 1) mod A007310(m + 1) is congruent to 0, otherwise 0.
Table begins
  \m  1 2  3  4  5  6  7  8  9  10  11  12  13  14  15  16 ...
  n\
   1| 5 0  0  0  0  0  0  0  0   0   0   0   0   0   0   0 ...
   2| 0 7  0  0  0  0  0  0  0   0   0   0   0   0   0   0 ...
   3| 0 0 11  0  0  0  0  0  0   0   0   0   0   0   0   0 ...
   4| 0 0  0 13  0  0  0  0  0   0   0   0   0   0   0   0 ...
   5| 0 0  0  0 17  0  0  0  0   0   0   0   0   0   0   0 ...
   6| 0 0  0  0  0 19  0  0  0   0   0   0   0   0   0   0 ...
   7| 0 0  0  0  0  0 23  0  0   0   0   0   0   0   0   0 ...
   8| 5 0  0  0  0  0  0 25  0   0   0   0   0   0   0   0 ...
   9| 0 0  0  0  0  0  0  0 29   0   0   0   0   0   0   0 ...
  10| 0 0  0  0  0  0  0  0  0  31   0   0   0   0   0   0 ...
  11| 5 7  0  0  0  0  0  0  0   0  35   0   0   0   0   0 ...
  12| 0 0  0  0  0  0  0  0  0   0   0  37   0   0   0   0 ...
  13| 0 0  0  0  0  0  0  0  0   0   0   0  41   0   0   0 ...
  14| 0 0  0  0  0  0  0  0  0   0   0   0   0  43   0   0 ...
  15| 0 0  0  0  0  0  0  0  0   0   0   0   0   0  47   0 ...
  16| 0 7  0  0  0  0  0  0  0   0   0   0   0   0   0  49 ...
For the n-th row of this square array, the leftmost terms, other than 0, are the factors of A(n,n). A(n,n) = A007310(n + 1). If for every m, m < n, A(m,n) = 0, then a(n) = A007310(n + 1) and A007310(n + 1) is prime.

G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 2, Section 2, Problems 96 and 105.

Davis Smith, Table of n, a(n) for n = 1..1000

Cf. A000034, A007310, A010729, A017281, A017365, A020639, A091999, A273669, A306277, A306285, A306331.

Cf. A107744, A111863 (bisections).

seq(min(op(numtheory[factorset] (6*ceil(n/2)+(-1)^n))), n=1..64) ;

FactorInteger[Rest@ Flatten@ Array[6 # + {1, 5} &, 33, 0]][[All, 1, 1]] (* Michael De Vlieger, Feb 15 2019 *)
FactorInteger[#][[1,1]]&/@Select[Range[2,200],CoprimeQ[#,6]&] (* Harvey P. Dale, Jul 10 2020 *)

for(n=2, 211, if((n%6==1)||(n%6==5), print1(factor(n)[1,1], ", ")))

vector(64,n,factor(6*ceil(n/2)+(-1)^n)[1,1])

a(n) = n++; factor(n\2*6-(-1)^n)[1,1]; \\ Michel Marcus, Feb 06 2019

a(n) = A020639(A007310(n + 1)).
a(n) = A020639(3n + A000034(n + 1)).
a(n) = A020639(6*ceiling(n/2) + (-1)^n).
a(floor(prime(n + 2)/3)) = prime(n + 2).

A322489 Numbers k such that k^k ends with 4.

Original entry on oeis.org

2, 18, 22, 38, 42, 58, 62, 78, 82, 98, 102, 118, 122, 138, 142, 158, 162, 178, 182, 198, 202, 218, 222, 238, 242, 258, 262, 278, 282, 298, 302, 318, 322, 338, 342, 358, 362, 378, 382, 398, 402, 418, 422, 438, 442, 458, 462, 478, 482, 498, 502, 518, 522, 538, 542, 558

Offset: 1

Bruno Berselli, Dec 12 2018

Also numbers k == 2 (mod 4) such that 2^k and k^2 end with the same digit.

Numbers congruent to {2, 18} mod 20. - Amiram Eldar, Feb 27 2023

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004526, A056849.
Subsequence of A139544, A235700.
Numbers k such that k^k ends with d: A008592 (d=0), A017281 (d=1), A067870 (d=3), this sequence (d=4), A017329 (d=5), A271346 (d=6), A322490 (d=7), A017377 (d=9).

GAP
```
List([1..70], n -> 10*n+3*(-1)^n-5);
```

[10*n+3*(-1)^n-5 for n in 1:70] |> println

Magma
```
[10*n+3*(-1)^n-5: n in [1..70]];
```

select(n->n^n mod 10=4,[$1..558]); # Paolo P. Lava, Dec 18 2018

Mathematica
```
Table[10 n + 3 (-1)^n - 5, {n, 1, 60}]
```
Maxima
```
makelist(10*n+3*(-1)^n-5, n, 1, 70);
```

apply(A322489(n)=10*n+3*(-1)^n-5, [1..70]) \\ M. F. Hasler, Dec 14 2018

Vec(2*x*(1 + 8*x + x^2) / ((1 - x)^2*(1 + x)) + O(x^70)) \\ Colin Barker, Dec 13 2018

[10*n+3*(-1)**n-5 for n in range(1, 70)]

Sage
```
[10*n+3*(-1)^n-5 for n in (1..70)]
```

O.g.f.: 2*x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 2 + 3*exp(-x) + 5*(2*x - 1)*exp(x).
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 10*n + 3*(-1)^n - 5. Therefore:
a(n) = 10*n - 8 for odd n;
a(n) = 10*n - 2 for even n.
a(n+2*k) = a(n) + 20*k.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(2*Pi/5)*Pi/20 = sqrt(5+2*sqrt(5))*Pi/20. - Amiram Eldar, Feb 27 2023

A346508 Positive integers k such that 10*k+1 is equal to the product of two integers greater than 1 and ending with 1 (A346507).

Original entry on oeis.org

12, 23, 34, 44, 45, 56, 65, 67, 78, 86, 89, 96, 100, 107, 111, 122, 127, 128, 133, 144, 149, 155, 158, 166, 168, 170, 177, 188, 189, 191, 199, 209, 210, 212, 220, 221, 232, 233, 243, 250, 251, 254, 260, 265, 275, 276, 282, 287, 291, 296, 298, 309, 311, 313, 317

Offset: 1

Stefano Spezia, Jul 21 2021

			107 is a term because 21*51 = 1071 = 107*10 + 1.

Stefano Spezia, Table of n, a(n) for n = 1..10000

Cf. A016873 (ending with 5), A017281, A324298 (ending with 6), A346507, A346509, A346510.

a={}; For[n=1, n<=350, n++, For[k=1, kMax[10a+1], AppendTo[a, n]]]]; a

def aupto(lim): return sorted(set(a*b//10 for a in range(11, 10*lim//11+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim))
print(aupto(318)) # Michael S. Branicky, Aug 21 2021

a(n) = (A346507(n) - 1)/10.
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
The conjecture is true since a(n) = (A346507(n) - 1)/10 and lim_{n->infinity} A346507(n)/A346507(n-1) = 1. - Stefano Spezia, Aug 21 2021

cp's OEIS Frontend

A169669 (first digit of n) * (last digit of n) in decimal representation.

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A306277 Numbers congruent to 1 or 8 mod 10.

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A348487 Positive numbers whose square starts and ends with exactly one 1.

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A154428 Primes of the form 50n^2 + 10n + 1.

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A168459 a(n) = (10*n + 5*(-1)^n - 3)/2.

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A208260 Nonprime numbers starting and ending with digit 1.

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A273372 Squares ending in digit 1.

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A168459 a(n) = (10n + 5(-1)^n - 3)/2.